Generalised differences and a class of multiplier operators in Fourier ...

arXiv:1503.04524v1 [math.CA] 16 Mar 2015

Generalised differences and a class of multiplier operators in Fourier analysis Rodney Nillsen School of Mathematics and Applied Statistics University of Wollongong, Northfields Avenue, Wollongong New South Wales, 2522, AUSTRALIA email address: [email protected]

Abstract Any zeros in the multiplier of an operator impose a condition on a function for it to be in the range of the operator. But if each function in a certain family F of functions satisfies such a condition, when is F the range of the operator? Let α, β ∈ Z, for g ∈ L2 ([0, 2π]) let gb be the sequence of Fourier coefficients of g, and let D denote differentiation. We consider the operator D2 − i(α + β)D − αβ on the second order Sobolev space of L2 ([0, 2π]). The multiplier of this operator is −(n − α)(n − β), so that b g(α) = gb(β) = 0 for any function g in the range of the operator. Let δx denote the Dirac measure at x, and let ∗ denote convolution. If b ∈ [0, 2π] let λb be the measure i i α−β α+β 1 h ib( α−β 1 h ib( α+β 2 ) + e−ib( 2 ) δ − 2 ) δ + e−ib( 2 ) δ e e 0 b −b . 2 2 A function of the form λb ∗ f is called a generalised difference, and we let F be the family of functions h such that h is some finite sum of generalised differences. It is shown that F is a closed subspace of L2 (T) that equals the range of D2 − i(α + β)D − αβ, and that every function in F is a sum of five generalised differences. The methods use partitions of [0, π/2] and estimates of integrals in Euclidean space. There are applications to the automatic continuity of linear forms.

2010 Mathematics Subject Classification. Primary 42A16, 42A45, Secondary 43A15, 46H40 Key words and phrases. Circle group, Fourier analysis, L2 spaces, multiplier operators, generalised differences, compact abelian groups, automatic continuity

1

Rodney Nillsen

1

Generalised differences and multiplier operators

Introduction

The circle group {z : |z| = 1} is denoted by T. The mapping x 7−→ eix from [0, 2π) to T means that we can and will identify T with [0, 2π) or [0, 2π] in the usual way, and both settings will be used (see the comments in [12, page 1034]). The group operation + on [0, 2π) is written additively and is the usual addition if 0 ≤ x + y < 2π, and is x + y − 2π if 2π ≤ x + y. The space L2 (T) is the Hilbert space of square integrable complex functions on T, and M (T) denotes the set of bounded, regular, complex Borel measures on T. Let Z denote the set of integers, and let N denote the set of positive integers. The convolution operation in M (T) is denoted by ∗. Thus, if µ ∈ M (T) and n ∈ N, µn denotes µ ∗ µ ∗ · · · ∗ µ, where µ appears n times. Considering n ∈ Z, f ∈ L2 (T) and µ ∈ M (T), we define the Fourier coefficients fb(n) and µ b(n) by 1 fb(n) = 2π

Z

2π

0

f (t) e−int dt and µ b(n) =

Z

2π

e−int dµ(t).

0

Note that for µ, ν ∈ M (T), (µ ∗ µ)b = µ b νb. Letting δx denote the Dirac measure at x, we see that δbx (n) = e−inx for x ∈ [0, 2π], and δbz (n) = z −n for z ∈ T. For s = 0, 1, 2, . . . the Sobolev space W s (T) is defined by ∞ o n X |n|2s |fb(n)|2 < ∞ . W s (T) = f : f ∈ L2 (T) and n=−∞

The differential operator D maps W 1 (T) into L2 (T) and can be defined by the property D(f )b(n) = infb(n), for all n ∈ Z.

We say that D is a multiplier operator on W 1 (T) with multiplier in. Note that this means that if f ∈ W 1 (T), D(f )b(0) = 0. Conversely, it is easy to see that if g ∈ L2 (T) and b g(0) = 0, then g = D(f ) for some f ∈ W 1 (T). Similarly, D s is a multiplier operator from W s (T) into L2 (T) with multiplier (in)s . Let α, β ∈ Z and s ∈ N be given, and let I denote the identity operator. In this paper, and for s ∈ N, we are concerned with operators (D 2 − i(α+ β)D − αβI)s that map W 2s (T) into L2 (T). These are multiplier operators with multipliers (−1)s (n−α)s (n−β)s . That is, for all f ∈ W s (T),  2  (D − i(α + β)D − αβI)s (f ) b(n) = (−1)s (n − α)s (n − β)s fb(n), 2

Rodney Nillsen

Generalised differences and multiplier operators

for all n ∈ Z. Note that the zeros of the multiplier of (D 2 −i(α+β)D−αβI)s are α and β. The operator (D 2 − i(α + β)D − αβI)s acting on f ∈ W 2s (T) eliminates the frequencies α and β from f . A particular case of the above is when α = −β. Then the operator (D 2 − i(α + β)D − αβI)s becomes (D 2 + α2 I)s , and it has the multiplier −n2 + α2 , which has zeros at −α and α. In order to give a feeling for the ideas in this paper, here is a statement of a special case of what is perhaps the main result. Let α ∈ Z. Then the following conditions (i), (ii) and (iii) are equivalent for a function f ∈ L2 (T). (i) fb(−α) = fb(α) = 0. (ii) f is a sum of five functions, for each one h of which there are z ∈ T and g ∈ L2 (T) such that h = (z α + z −α )g − (δz + δz −1 ) ∗ g.

(1.1)

(iii) There is g ∈ W 2 (T) such that (D 2 + α2 I)(g) = f . Now a function of the form g − δz ∗ g might be called a first order difference, and one of the form (δ1 − δz )s ∗ g might be called a difference of order s (see [8, 9]). So, a function as in (1.1) we call a type of generalised difference. The form of generalised differences is derived from the factors of the multiplier of the operator, rather than by replacing the elements D and D 2 in the operator by first and second order differences. The present ideas in proving a result of the above type involve looking at the structure of partitions of [0, π/2] associated with the zeros in [0, π/2] of the functions sin((n − α)x) and sin((n − β)x), and estimating integrals in Rm over sets that are Cartesian products of sets in a refined partition. The results here are related to work of Meisters and Schmidt [5], where they proved the following result. The following conditions (i) and (ii) are equivalent for a function f ∈ L2 (T). (i) fb(0) = 0. (ii) f is a sum of three functions, for each one h of which there are z ∈ T and g ∈ L2 (T) such that h = g − δz ∗ g. They deduced from this that if T is a linear form on L2 (T) such that T (f ) = T (δz ∗ f ) for all z ∈ T and f ∈ L2 (T), then T is continuous. That is, any translation invariant linear form on L2 (T) is automatically continuous, and this was proved more generally for compact, connected abelian groups. Further work relating to the original results of Meisters and Schmidt can be found, for example, in [1, 4, 6, 7, 8, 9] where there are further references. One way to think of the ideas in [5] is that they are 3

Rodney Nillsen

Generalised differences and multiplier operators

concerned with the range of the differential operator D whose multiplier is a linear polynomial. On the other hand, the present work is concerned with differential operators whose multipliers are quadratic in nature. The work here goes back to [5], but takes some of the ideas there in a different direction from the mainstream of later work. There are also applications to automatic continuity of linear forms. A suitable general reference on classical Fourier series is [2], and for abstract harmonic analysis on groups see [3, 12].

2

Background and statement of the main result

P∞ There will be cause to consider series of the P form j=−∞ aj /bj , where aj , bj ≥ 0 for all j. In such a case, we write ∞ a j=−∞ j /bj = ∞ if there is a term of the form aj /0 with aj > 0. If there are any terms of the form 0/0 we eitherP neglect them in the sum or make them equal to 0. We P∞ ∞ write respectively a /b < ∞ or j=−∞ j j j=−∞ aj /bj = ∞ if the series P∞ j=−∞,aj >0 aj /bj converges or diverges in the usual sense. The following result is due essentially to Meisters and Schmidt [5, page 413]. Theorem 2.1 Let f ∈ L2 ([0, 2π]) and let µ1 , µ2 , . . . , µr ∈ M ([0, 2π]). Then the following conditions (i) and (ii) are equivalent. P (i) There are f1 , f2 , . . . , fr ∈ L2 ([0, 2π]) such that f = rj=1 µj ∗ fj . (ii)

∞ X

n=−∞

|fb(n)|2 < ∞. r X 2 |b uj (n)| j=1

Proof. This is essentially proved in [5, pages 411-412 ], but see also [8, pages 77-88] and [9, page 23]. A more accessible proof for the present context is on the world wide web [10].  In [5], the measures in Theorem 2.1 were taken to be of the form µb = δ0 − δb , in which case µb ([0, 2π]) = 0 and µbb (n) = 1 − e−ibn . In the context here, we let α, β ∈ Z and we will apply Theorem 2.1 using measures λb , b ∈ [0, 2π), where i i α−β α+β 1 h ib( α+β ) 1 h ib( α−β ) 2 2 e e + e−ib( 2 ) δ0 − δb + e−ib( 2 ) δ−b . (2.1) λb = 2 2 b Note the the Fourier transform λb of λb is given for n ∈ Z by       α−β α+β b λb (n) = cos b − cos n− b . (2.2) 2 2 4

Rodney Nillsen

Consequently,

Generalised differences and multiplier operators

λbb (α) = λbb (β) = 0, for all b ∈ [0, 2π].

(2.3)

So if b ∈ [0, 2π] and g ∈ L2 ([0, 2π]), i i α−β α+β 1 h ib( α+β ) 1 h ib( α−β ) 2 2 e e + e−ib( 2 ) g − δb + e−ib( 2 ) δ−b ∗ g, λb ∗ g = 2 2 (2.4) and we see that if f ∈ L2 ([0, 2π]) is a function of the form λb ∗ g, then fb(α) = fb(β) = 0. A function of the form λb ∗ g, as in (2.4) above, is called a generalised difference in L2 ([0, 2π]). The following is a crucial technical result. It is not proved here, but is a consequence of results in subsequent sections. Lemma 2.2 Let α, β ∈ Z and let s ∈ N be given. Then there is M > 0 such that for all n ∈ Z with n ∈ / {α, β}, Z dx1 dx2 · · · dxm       2s ≤ M. 4s+1 4s+1 X [0,2π] α − β α + β cos n− xj − cos xj 2 2 j=1

The central result proved here using Lemma 2.2 is the following.

Theorem 2.3 Let α, β ∈ Z. Then the following conditions (i), (ii), (iii) and (iv) on a function f ∈ L2 ([0, 2π]) are equivalent. (i) fb(α) = fb(β) = 0. (ii) There are m, s ∈ N, b1 , b2 , . . . , bm ∈ [0, 2π] and f1 , f2 , . . . , fm ∈ 2 L ([0, 2π]) such that f is equal to m h   is X α+β α−β α+β α−β ∗ fj . eibj ( 2 ) + e−ibj ( 2 ) δ0 − eibj ( 2 ) δbj + e−ibj ( 2 ) δ−bj j=1

(2.5) (iii) There are s ∈ N, b1 , b2 , . . . , b4s+1 ∈ [0, 2π] and f1 , f2 , . . . , f4s+1 ∈ L2 ([0, 2π]) such that f is equal to 4s+1 X h j=1

eibj (

α−β 2

  is ibj ( α+β −ibj ( α+β ) + e−ibj ( α−β 2 ) δ 2 )δ 2 )δ ∗ fj . 0 − e bj + e −bj (2.6)

(iv) There are s ∈ N and g ∈ W 2s ([0, 2π]) such that
D 2 − i(α + β)D − αβI (g) = f. 5

(2.7)

Rodney Nillsen

Generalised differences and multiplier operators

When the equivalent conditions (i), (ii), (iii) and (iv) are satisfied and s ∈ N is given, we have that for almost all (b1 , b2 , . . . , b4s+1 ) ∈ [0, 2π]2s+1 , there are f1 , f2 , . . . , f4s+1 ∈ L2 ([0, 2π]) such that (2.6) holds. Also, the functions in L2 ([0, 2π]) that can be written in the form (2.6) form a closed vector subspace of L2 ([0, 2π]). Proof. It is obvious that (iii) implies (ii). Now, (ii) implies (i) since, as noted in (2.3), the Fourier coefficients of any function appearing within the sum in (2.5) vanish at α and β. In order to prove that (i) implies (iii), let fb(α) = fb(β) = 0, and let M be the constant as in Lemma 2.2. If we integrate the function that maps (x1 , x2 , . . . , x4s+1 ) ∈ R4s+1 into X

n=−∞ n6=α,β

|fb(n)|2      2s  4s+1 X α−β α+β cos x x − cos n − j j 2 2

(2.8)

j=1

over [0, 2π]4s+1 , and interchange the order of integration and summation we obtain  

  ∞ Z  X dx1 dx2 · · · dx4s+1  b   |f (n)|2        4s+1 2s   4s+1 X [0,2π] α−β α+β n=−∞  cos n6=α,β xj − cos xj  n− 2 2 j=1

≤M

∞ X

n=−∞

< ∞.

|fb(n)|2

We deduce that for almost all (x1 , x2 , . . . , x4s+1 ) ∈ [0, 2π]4s+1 , the sum in (2.8) is finite. Using (2.2), we see that for almost all (x1 , . . . , x4s+1 ) ∈ [0, 2π]4s+1 , ∞ X

|fb(n)|2 4s+1 X n=−∞ s (n)|2 |λc xj j=1

6

Rodney Nillsen

=

∞ X

n=−∞

Generalised differences and multiplier operators fb(n)|2

4s+1 X j=1

=

∞ X

n=−∞

fb(n)|2

4s+1 X j=1

=

∞ X

n=−∞ n6=α,β

bx (n)|2s |λ j bx (n)|2s |λ j

|fb(n)|2      2s  4s+1 X α−β α+β cos x xj − cos n − j 2 2 j=1

< ∞.

It now follows from (2.2), and the equivalence of (i) and (ii) in Theorem 2.1, that for almost all (b1 , b2 , . . . b4s+1 ) ∈ [0, 2π]4s+1 there are f1 , f2 , . . . , f4s+1 ∈ L2 ([0, 2π]) such that 4s+1 X λsbj ∗ fj . f= j=1

We see, using (2.1), that (2.6) and hence (iii) hold. Now to see that (iv) implies (i), we observe that the multiplier of (D 2 − i(α + β)D − αβI)s is (−1)s (n − α)s (n − β)s . Thus, if f = (D 2 − i(α + β)D − αβI)s (g), as in (2.7), fb(α) = fb(β) = 0. Now, we prove (i) implies (iv). If fb(α) = fb(β) = 0, define g in terms of gb by putting gb(α) = b g(β) g (n) = (−1)s fb(n)/(n−α)s (n−β)s , for n 6= P∞= 0 and4sb α and n 6= β. Then n=−∞ |n| |b g (n)|2 < ∞, so g ∈ W 2s ([0, 2π]). Also, as 2 the multiplier of (D − i(α + β)D − αβI)s is (−1)s (n − α)s (n − β)s , we see that (D 2 −i(α+β)D−αβI)s (g)b = fb and so (D 2 −i(α+β)D−αβI)s (g) = f . Finally, that the functions expressible as in (2.6) form a closed subspace follows from the fact that such functions are characterised by (i). 

3

Partitioning, zeros and inequalities

In this section we make observations and obtain results for the later proving of Lemma 2.2.

7

Rodney Nillsen

Generalised differences and multiplier operators

Lemma 3.1 Let m, s ∈ N and α, β, n ∈ Z. Then, Z dx1 dx2 · · · dxm       2s m m X α−β α+β [0,2π] cos n− xj − cos xj 2 2 j=1 Z dx1 dx2 · · · dxm , = 22m−2s m m X [0,π/2] 2s 2s sin ((n − α)xj ) sin ((n − β)xj )

(3.1)

j=1

but note that both integrals may be infinite. Proof. The use of a familiar trigonometric formula and a simple substitution gives Z dx1 dx2 · · · dxm     2s   m α−β α+β [0,2π]m X cos xj − cos xj n− 2 2 j=1 Z dx1 dx2 · · · dxm . (3.2) = 2m−2s m X [0,π]m 2s 2s sin ((n − α)xj ) sin ((n − β)xj ) j=1

Note that for x ∈ R and ℓ ∈ Z , | sin(ℓx)| = | sin(ℓ(π − x))|. Also, note Q that [0, π)m is the disjoint union of the 2m sets m J t=1 t where, for each t, Jt is either [0, π/2) or [π/2, π). Using substitutions for π − xj , as needed in (3.2), we see that (3.1) follows from (3.2).  We are aiming to estimate the integral in Lemma 3.1. Motivated by (3.1), we consider the zeros in [0, π/2] of sin((n − α)x) and sin((n − β)x). Some preliminary notions are needed. Definitions. If J is an interval we denote its length by λ(J). Let [a, b] be a closed interval with λ([a, b]) > 0. A family {J0 , J1 , . . . , Jr−1 } of closed r−1 intervals having non-empty interiors is a partition of [a, b] if ∪j=0 Jj = [a, b] and any two intervals in the family have at most a single point in common. In such a case, the intervals may be arranged so that the right endpoint of Jj−1 is the left endpoint of Jj for all j = 1, 2, . . . , r − 1. Note that in the sense used here, the sets in a partition are not pairwise disjoint. Lemma 3.2 Let [a, b] be a closed interval with λ([a, b]) > 0. Let R0 , R1 , . . . , Rr−1 be closed intervals in a partition P1 of [a, b]. Let S0 , S1 , . . . , Ss−1 be 8

Rodney Nillsen

Generalised differences and multiplier operators

closed intervals in a partition P2 of [a, b]. Let  A = (j, k) : 0 ≤ j ≤ r − 1, 0 ≤ k ≤ s − 1 and λ(Rj ∩ Sk ) > 0 , and put

 P = Rj ∩ Sk : (j, k) ∈ A .

(3.3)

Then, P is a partition of [a, b] and the number of intervals in P is at most r + s − 1.

(3.4)

Proof. It is clear that P = {Rj ∩Sk : (j, k) ∈ A} as in (3.3) is a partition of [a, b]. Given any two partitions P1 , P2 as in the Lemma, we will denote the partition given as in (3.3) by P(P1 , P2 ). We proceed by induction on r. If r = 1, P1 = {R0 } = {[a, b]}, so that P(P1 , P2 ) = P2 and P(P1 , P2 ) has s elements. In this case, s = r + s − 1, and we see that when r = 1, (3.4) holds for all s ∈ N. Similarly, it is easy to see that when r = 2, (3.4) holds for all s ∈ N. So, let r ∈ N with r ≥ 3 be such that (3.4) holds for all partitions P1 having r intervals and for all partitions P2 having any number of intervals. Let P3 = {Q0 , Q1 , . . . , Qr } be a partition of [a, b] into r + 1 intervals and let P2 = {S0 , S1 , . . . , Ss−1 } be a partition of [a, b] into s intervals. Consider the partition P1 = {Q0 ∪ Q1 , Q2 . . . , Qr }, which has r intervals. By the inductive hypothesis, P(P1 , P2 ) has at most r + s − 1 intervals. Let ξ be the right endpoint of Q0 , which is also the left endpoint of Q1 . Now, in passing from P1 to P3 , we do so by using ξ to divide the single interval Q0 ∪ Q1 into the two intervals Q0 and Q1 . If ξ is not an endpoint of any interval in P2 , this will divide one interval in P(P1 , P2 ) into two subintervals belonging to P(P2 , P3 ). If ξ is an endpoint of some interval in P2 , this division does not increase the number of intervals in P(P2 , P3 ) in going from P1 to P3 . In either case, we see that P(P2 , P3 ) has at most r + s − 1 + 1 = r + s = (r + 1) + s − 1 intervals, showing that (3.4) holds with r + 1 in place of r. The result follows by induction.  Definition. Let [a, b] be a closed interval of positive length. Let P1 = {R0 , R1 , . . . , Rr−1 } and P2 = {S0 , S1 , . . . , Ss−1 } be two partitions of [a, b]. Let P be the partition of [a, b] as given by (3.3) in Lemma 3.2. Then P is called the refinement of the partitions P1 and P2 . 9

Rodney Nillsen

Generalised differences and multiplier operators

Now, let n, γ ∈ Z with n 6= γ be given. We construct an associated partition P(γ) of [0, π/2], as follows. (i) When |n − γ| is even we define (|n − γ| + 2)/2 closed subintervals Q0 , Q1 , . . . , Q|n−γ|/2 of [0, π/2] by putting     π(|n − γ| − 1) π π , , Q|n−γ|/2 = , and Q0 = 0, 2|n − γ| 2|n − γ| 2   π(j − 1/2) π(j + 1/2) Qj = , , (3.5) |n − γ| |n − γ|

for j = 1, 2 . . . , (|n − γ| − 2)/2. (ii) When |n − γ| is odd we define (|n − γ| + 1)/2 closed subintervals Q0 , Q1 , . . . , Q(|n−γ|−1)/2 of [0, π/2] by putting     π(j − 1/2) π(j + 1/2) π , and Qj = , (3.6) Q0 = 0, 2|n − γ| |n − γ| |n − γ|

the latter for j = 1, 2 . . . , (|n − γ| − 1)/2. Put θ(r) = (r + 2)/2 if r is even, and θ(r) = (r + 1)/2 if r is odd. Then, with n and γ as given, (3.5) and (3.6) above define θ(|n − γ|) closed subintervals Q0 , Q1 , . . . , Qθ(|n−γ|)−1 of [0, π/2]. We put  P(γ) = Q0 , Q1 , Q2 , . . . , Qθ(|n−γ|)−1 . (3.7)

Note that P(γ) depends only upon |n − γ| so, strictly speaking, P(γ) depends also on n as well as γ. The significance of P(γ) lies in its relationship to the zeros of sin(n − γ)x in [0, π/2]. There are θ(|n − γ|) zeros c1 , c2 , . . . , cθ(|n−γ|) of sin(n − γ)x in [0, π/2] given by cj =

πj , for j = 0, 1, 2, . . . , θ(|n − γ|) − 1. |n − γ|

(3.8)

Now, we see from (3.5) and (3.6) that c0 is the left endpoint of Q0 , when |n − γ| is even cj is the midpoint of Qj for j = 1, 2 . . . , θ(|n − γ|) − 2 and cθ(|n−γ|)−1 = π/2 is the right endpoint of Qθ(|n−γ|)−1 and, when |n − γ| is odd cj is the midpoint of Qj for j = 1, 2 . . . , θ(|n − γ|) − 1. Lemma 3.3 Let α, β, n ∈ Z be such that n 6= α and n 6= β. Let P(α, β) be the partition of [0, π/2] that is the refinement of P(α) and P(β), as given by (3.3). Then the number of intervals in P(α, β) is bounded above by  2 max |n − α|, |n − β| . (3.9) 10

Rodney Nillsen

Generalised differences and multiplier operators

Also, if J ∈ P(α, β), 0 < λ(J) ≤ min



π π , |n − α| |n − β|



.

(3.10)

Proof. The partition P(α) has θ(|n − α|) intervals, while P(β) has θ(|n − β|) intervals. So, we see from (3.4) that P(α, β) has at most θ(|n − α|) + θ(|n − β|) − 1 intervals. However, θ(r) ≤ (r + 2)/2 for all r ∈ N, so an upper bound for the number of intervals in P(α, β) is 1 (|n − α| + |n − β|) + 1 ≤ max{|n − α|, |n − β|} + 1 ≤ 2 max{|n − α|, |n − β|}. 2 Finally, if J ∈ P(α, β), J = R ∩ S for some R ∈ P(α) and S ∈ P(β). Then, λ(R) ≤ π/|n − α| and λ(S) ≤ π/|n − β|, and so (3.10) follows.  Figure 1 illustrates Lemma 3.3 in the case α = 1, β = −1 and n = 9. π/4

0

a 0 = b0

b1 a1

b2

a2

π/2

b3

a3 b4

a4 = b5

Figure 1. The figure illustrates the case n = 9, α = 1, β = −1. We have n − α = 8, n − β = 10. The zeros of sin 8x and sin 10x in [0, π/2] are denoted respectively by aj and bj . The vertical lines in the figure illustrate the intervals in the refinement P(−1, 1) of P(−1) and P(1). Note that four of the ten intervals in P(−1, 1) contain no zeros of sin 8x sin 10x.

Now, if x ∈ R, let dZ (x) denote the distance from x to a nearest integer. For later use, note the fact that dZ (x) = |x| if and only if −1/2 ≤ x ≤ 1/2.

Lemma 3.4 Let n, α, β ∈ Z with n 6= α, n 6= β. Let the partitions P(α) and P(β) be given as in (3.7), and we write P(α) = {R0 , R1 , . . . , Rθ(|n−α|)−1 } and P(β) = {S0 , S1 , . . . , Sθ(|n−β|)−1 }.

Let Rj ∩ Sk be an element of the refinement P(α, β) of P(α) and P(β). Then, if x ∈ Rj ∩ Sk , we have sin2 ((n − α)x) sin2 ((n − β)x)  2  2 jπ 24 (n − α)2 (n − β|2 kπ x − . x − ≥ π4 |n − α| |n − β| 11

(3.11)

Rodney Nillsen

Generalised differences and multiplier operators

Proof. We will use the fact that for all x ∈ R, | sin πx| ≥ 2dZ (x) [9, page 89, for example]. Given that x ∈ Rj , we see from (3.5) and (3.6) that   1 j ≤ . (n − α) x − (3.12) π |n − α| 2 Using (3.12) we now have, for all x in Rj ,

| sin((n − α)x)| = | sin(|n − α|x − jπ)|    j x − = sin π(n − α) π |n − α|    x j ≥ 2dZ (n − α) − π |n − α| x j = 2|n − α| − π |n − α| 2 jπ = |n − α| x − . π |n − α|

Using a corresponding argument, we see also that for all x ∈ Sk , kπ 2 . | sin((n − β)x)| ≥ |n − β| x − π |n − β| Conclusion (3.11) now follows from (3.13) and (3.14).

4

(3.13)

(3.14) 

Integral estimates in Rm

In this section we develop estimates for some integrals in Rm , and an inequality between quadratics, with a view to proving Lemma 2.2. Lemma 4.1 Let s, m ∈ N with m ≥ 4s + 1. Then, there is a number M > 0, depending upon s andQm only, such that for all b1 , b2 , . . . , bm > 0 and for all (a1 , a2 , . . . , am ) ∈ m t=1 [−bt , bt ], Z om−4s  n du1 du2 . . . dum ≤ M max b , b , . . . , b . 1 2 m Qm m X 2 2 2s t=1 [−bt , bt ] (ut − at ) t=1

12

Rodney Nillsen

Generalised differences and multiplier operators

PROOF. Clearly, we may assume that 0 ≤ at ≤ bt for all t = 1, 2, . . . , m. Now, we have Z du1 du2 . . . dum Qm m X t=1 [−bt , bt ] (u2t − a2t )2s Z

t=1

du1 du2 . . . dum m X t=1 [0, bt ] (u2t − a2t )2s

= 2m Qm Z

t=1

du1 du2 . . . dum

= 2m Qm Z

t=1 [0, bt ]

m X t=1

(ut − at )2s (ut + at )2s

dv1 dv2 . . . dvm , m X 2s 2s j=1 [−at , bt − at ] vt (vt + 2at )

= 2m Qm

(4.1)

t=1

on putting vt = ut − at . Now observe that if vt ≥ 0 then vt + 2at ≥ 0, and that if −at ≤ vt ≤ 0 then vt + 2at ≥ at ≥ |vt |. Also, there is Cm > 0 such that for all (v1 , v2 , . . . , vm ) ∈ Rm , m X

vj4s

j=1

2s  m X ≥ Cm  vj2  .

(4.2)

j=1

If r > 0, we denote the closed sphere {x : x ∈ Rm and |x| ≤ r} by S(0, r). Also, we put b = max{b1 , b2 , . . . , bm }. Using the preceding observations and (4.1) we have Z du1 du2 . . . dum Qm m X t=1 [−bt , bt ] (u2t − a2t )2s Z

t=1

dv1 dv2 . . . dvm m X j=1 [−at , bt − at ] vt4s

≤ 2m Qm

t=1

13

Rodney Nillsen

Generalised differences and multiplier operators

Z dv1 dv2 . . . dvm 2m ≤ !2s , using (4.2), Qm m Cm X j=1 [−at , bt − at ] 2 vt t=1 Z dv1 dv2 . . . dvm 2m ≤ !2s , Qm m Cm X j=1 [−bt , bt ] vt2 t=1

as [−at , bt − at ] ⊆ [−b, b]m ,

2m ≤ Cm

=

Z

√ S(0,b m)

2m+1 π m/2 Cm Γ(m/2)

Z

√ dv1 dv2 . . . dvm , as [−b, b]m ⊆ S(0, b m), ! 2s m X vt2

√ b m

t=1

r m−4s−1 dr, by [14, pages 394-395],

0

2m+1 π m/2 m(m−4s)/2 bm−4s = . Cm Γ(m/2)(m − 4s) −1 2m+1 π m/2 m(m−4s)/2 (m − 4s)−1 Γ(m/2)−1 . So, the result holds if M = Cm 

Lemma 4.2 Let s, m ∈ N with m ≥ 4s + 1, and let numbers bt,k , ct and dt be given for all t, k with t = 1, 2 . . . , m and k = 1, 2. We assume that 0 ≤ bt,1 ≤ ct ≤ dt ≤ bt,2 for all t = 1, 2, . . . , m. Then, there is a number M > 0, depending upon s and m only and independent of bt,k , ct and dt , such that Z du1 du2 . . . dum Qm m X t=1 [bt,1 , bt,2 ] (ut − ct )2s (ut − dt )2s t=1

≤M



om−4s n max b1,2 − b1,1 , b2,2 − b2,1 , . . . , bm,2 − bm,1 .

Proof. Put, for t = 1, 2, . . . , m, ηt =

ct + dt dt − ct , γt = , vt = ut − ηt . 2 2 14

Rodney Nillsen

Generalised differences and multiplier operators

Note that ηt ∈ [bt,1 , bt,2 ] and 0 ≤ γt ≤ (bt,2 − bt,1 )/2. Using Lemma 4.1 and putting vt = ut − ηt in the following we have Z du1 du2 . . . dum Qm m X t=1 [bt,1 , bt,2 ] (ut − ct )2s (ut − dt )2s t=1

Z

= Qm

t=1

Z

dv1 dv2 . . . dvm

m [bt,1 − ηt , bt,2 − ηt ] X

= Qm Z

t=1

≤ Qm

t=1

t=1

(vt + γt )2s (vt − γt )2s

dv1 dv2 . . . dvm m
2s [−(ηt − bt,1 ), bt,2 − ηt ] X 2 vt − γt2 t=1

dv1 dv2 . . . dvm , m X
[−(bt,2 − bt,1 ), bt,2 − bt,1 ] 2 2 2s vt − γt t=1

as [−(ηt − bt,1 ), bt,2 − ηt ] ⊆ [−(bt,2 − bt,1 ), bt,2 − bt,1 ]. We now see that Lemma 4.1 applies because 0 ≤ γt ≤ bt,2 − bt,1 , and so there is a constant M > 0, depending only upon s and m, such that Z du1 du2 . . . dum Qm m X t=1 [bt,1 , bt,2 ] (ut − ct )2s (ut − dt )2s t=1

om−4s  n . ≤ M max b1,2 − b1,1 , b2,2 − b2,1 , . . . , bm,2 − bm,1

Lemma 4.3 Let c ≤ a < b ≤ d. Let f , g be the quadratic functions given by f (x) = (x − c)(d − x), g(x) = (x − a)(b − x). Then, f (x) ≥ g(x) ≥ 0 for all a ≤ x ≤ b. Proof. We have (f − g)(x) = x(c + d − a − b) + ab − cd. So, (f − g)(a) = (a − c)(d − a) ≥ 0 and (f − g)(b) = (d − b)(b − c) ≥ 0. As f − g is linear and non-negative at a and b, we deduce that f (x) ≥ g(x) for all x ∈ [a, b].  15



Rodney Nillsen

5

Generalised differences and multiplier operators

Completion of the proof of Theorem 2.3

Let s ∈ N and n, α, β ∈ Z with n 6= α and n 6= β. Let a0 , a1 , . . . , aθ(|n−α|)−1 and b0 , b1 , . . . , bθ(|n−β|)−1 respectively be the zeros of sin(n−α)x and sin(n− β)x in [0, π/2], as given correspondingly for sin(n − γ)x in (3.8). Let P(α) = {R0 , R1 , . . . , Rθ(|n−α|)−1 } and P(β) = {S0 , S1 , . . . , Sθ(|n−β|)−1 } be the partitions as given by (3.7), and recall that A is the set of all (j, k) such that λ(Rj ∩ Sk ) > 0 and that the partition P(α, β) of [0, π/2] is the set {Rj ∩ Sk : (j, k) ∈ A}, as in Lemma 3.3. We see from (3.1) of Lemma 3.1 and from (3.11) of Lemma 3.4 that for any m, s ∈ N, Z dx1 dx2 · · · dxm       2s m m X (α − β) (α + β) [0,2π) cos n− xj − cos xj 2 2 j=1 Z dx1 dx2 · · · dxm = 22m−2s m m X [0,π/2) sin2s ((n − α)xj ) sin2s ((n − β)xj ) j=1

≤

22m−6s π 4s

(n −

α)2s (n

−

β)2s

X

(j1 ,k1 ),...,(jm ,km )∈A


J (j1 , k1 ), . . . , (jm , km ) ,

(5.3)

where Z
J (j1 , k1 ), . . . , (jm , km ) = Qm

t=1 Rjt

∩ Skt

m X t=1

dx1 dx2 · · · dxm 2s

(xt − ajt ) (xt − bkt )

. 2s

(5.4) Note that by Lemma 3.3, the number of terms in the sum in (5.3) is bounded by 2m max{|n − α|m , |n − β|m }. (5.5) Now, let (j1 , k1 ), . . . , (jm , km ) ∈ A, and let t ∈ {1, 2, . . . , m}. We consider the following possibilites (i), (ii) and (iii). (i) If ajt ∈ Rjt ∩ Skt and bkt ∈ Rjt ∩ Skt we put a′jt = ajt and b′kt = bkt . Note that if ajt = bkt , then both ajt and bkt belong to Rjt ∩ Skt . (ii) If ajt ∈ Rjt ∩ Skt but bkt ∈ / Rjt ∩ Skt , we put a′jt = ajt and take b′kt to be the endpoint of Rjt ∩ Skt that is closest to bkt . If ajt ∈ / Rjt ∩ Skt but bkt ∈ Rjt ∩ Skt , we take a′jt to be the endpoint of Rjt ∩ Skt that is closest to ajt , and we put b′kt = bkt . 16

Rodney Nillsen

Generalised differences and multiplier operators

(iii) If ajt ∈ / Rjt ∩ Skt , it must happen that ajt lies / Rjt ∩ Skt and bkt ∈ to the left of Rjt ∩ Skt and bkt to the right, or vice versa. In either case, we put a′jt to be one endpoint of Rjt ∩ Skt and b′kt to be the other. Now it is clear that in the cases (i) and (ii) above, for all xt ∈ Rjt ∩ Skt we have (5.6) |xt − ajt |2s |xt − bkt |2s ≥ |xt − a′jt |2s |xt − b′kt |2s . In case (iii) above, we see from the simple result on quadratics in Lemma 4.3 that (5.6) holds for all xt ∈ Rjt ∩ Skt . All possibilities are exhausted by (i), (ii) and (iii), so that for all jt , kt we see that a′jt , b′kt ∈ Rjt ∩ Skt and that (5.6) holds for all xt ∈ Rjt ∩ Skt . Now assume that m ∈ N with m ≥ 4s + 1. We have from (5.4) and (5.6) that there is M > 0, depending on s and m only, such that
J (j1 , k1 ), . . . , (jm , km ) Z dx1 dx2 · · · dxm = Qm m X t=1 Rjt ∩ Skt |xt − ajt |2s |xt − bkt |2s Z

t=1

≤ Qm

t=1 Rjt

∩ Skt

m X t=1

dx1 dx2 · · · dxm 2s

, by (5.6), 2s

|xt − ajt′ | |xt − bkt′ |


m−4s , ≤ M max λ(Rj1 ∩ Sk1 ), λ(Rj2 ∩ Sk2 ), . . . , λ(Rjm ∩ Skm ) 

1 1 , |n − α|m−4s |n − β|m−4s π m−4s M = . max{|n − α|m−4s , |n − β|m−4s } ≤ π m−4s M min



by Lemma 4.2,

, using (3.10),

Now, using (5.5), we see from (5.3) and (5.7) that Z dx1 dx2 · · · dxm     2s   m (α − β) (α + β) [0,2π)m X cos x xj − cos n − j 2 2 ≤

j=1 3m−6s 2 πm M

(n − α)2s (n − β)2s

·

max{|n − α|m , |n − β|m } max{|n − α|m−4s , |n − β|m−4s }

17

(5.7)

Rodney Nillsen

Generalised differences and multiplier operators

3m−6s m

=2

π M · max

≤ 23m−6s π m M K,



(n − α)2s (n − β)2s , (n − β)2s (n − α)2s



(5.8)

where K > 0 is a suitable constant chosen to be independent of n. Lemma 2.2 is immediate from (5.8) upon taking m to be 4s + 1 and, as discussed in Section 2, Theorem 2.3 is now established. 

6

A sharpness result

If was shown in Theorem 2.3 that if f ∈ L2 ([0, 2π]) is such that fb(α) = fb(β) = 0, then for almost all (b1 , b2 , . . . , b4s+1 ) ∈ [0, 2π]4s+1 , f can be written in the form (2.6) and consequently in the form (2.5). However, in this section we show that if m ∈ N and b1 , b2 , . . . , bm ∈ [0, 2π]4s+1 are given, there are many functions with fb(α) = fb(β) = 0 that cannot be written in the form (2.5). Thus, no single choice of b1 , b2 , . . . , bm suffices to ensure that (2.5) is possible for all f ∈ L2 ([0, 2π]) such that fb(α) = fb(β) = 0. The methods extend the techniques in [5, pages 420-421]. Lemma 6.1 Let c1 , c2 , . . . , cm ∈ R. Then, there are infinitely many q ∈ N such that dZ (qcj ) < 1/q 1/m for all j = 1, 2, . . . , m. Proof. See [11, Theorem 4.6] or [13, page 27], for example.



Theorem 6.2 Let m, s ∈ N and let α, β ∈ Z be given. Also, let c1 , c2 , . . . , cm ∈ [0, 2π] be given. Then, there is a vector subspace V of L2 ([0, 2π]) such that V has algebraic dimension equal to that of the continuum but, for any f ∈ V with f 6= 0, there is no choice of f1 , f2 , . . . , fm ∈ L2 ([0, 2π]) such that f is equal to m h   is X α+β α−β α+β α−β ∗ fj . eicj ( 2 ) + e−icj ( 2 ) δ0 − eicj ( 2 ) δcj + e−icj ( 2 ) δ−cj j=1

Proof. Let f1 , f2 , . . . , fm ∈ L2 ([0, 2π]), for b ∈ [0, 2π] let λb be given by (2.1), and let f be given by f=

m X j=1

λscj ∗ fj .

18

(6.9)

Rodney Nillsen

Generalised differences and multiplier operators

Then, using (2.2), for all n ∈ Z we have fb(n) =

m X

bc (n)s fbj (n) λ j

j=1 m  X

   s  α+β cj − cos cj fbj (n) n− = cos 2 j=1     m X s (n − β)cj s (n − α)cj s fbj (n). sin sin =2 2 2 

α−β 2

j=1

Thus, as | sin x| ≤ πdZ (x/π), we see that |fb(n+α)| ≤ 2s

m m  nc  s  nc s X X j j s s b dZ |fbj (n+α)|. sin |fj (n+α)| ≤ π 2 2 2π j=1

j=1

(6.10) Now, by Lemma 6.1, for each ℓ ∈ N and k ∈ {1, 2 . . . , 2ℓ } there is qk,ℓ ∈ Z such that   c  1 j < 1/m , for all j = 1, 2, . . . , m. (6.11) dZ qk,ℓ 2π ℓ

We see also from Lemma 6.1 that the integers qk,ℓ can be chosen so that they are all distinct, over all ℓ ∈ N and k ∈ {1, 2, . . . , 2ℓ }. Now, let Φ be the family of all functions φ : N 7−→ N such that φ(1) ∈ {1, 2} and φ(ℓ + 1) ∈ {2φ(ℓ) − 1, 2φ(ℓ)} for all ℓ ∈ N. Note that if φ, φ′ ∈ Φ and φ(j) 6= φ′ (j), then φ(n) 6= φ′ (n), for all n > j. (6.12)

We see that φ(n) ∈ {1, 2, . . . , 2n } for all n ∈ N, and that Φ has the cardinality of the continuum. Now, if φ ∈ Φ, define a function fφ as follows. If n ∈ Z and n = qφ(ℓ),ℓ + α for some ℓ ∈ N, then ℓ is unique and we put fbφ (n) =

1 ℓ1/2+s/m

.

If n ∈ / {qφ(ℓ),ℓ + α : ℓ ∈ N}, we put Then, because

∞ X

n=−∞

fbφ (n) = 0. |fbφ (n)|2 =

∞ X ℓ=1

19

1

ℓ1+2s/m

< ∞,

Rodney Nillsen

Generalised differences and multiplier operators

we see that fφ ∈ L2 ([0, 2π]). Now, assume that φ ∈ Φ and that fφ can be put in the form (6.9). By (6.10) and (6.11) and taking n = qφ(ℓ),ℓ where ℓ ∈ N we have |fbφ (qφ(ℓ),ℓ +α)| ≤

π s 2s ℓs/m

 

m X j=1



|fbj (qφ(ℓ),ℓ + α)| ≤

Cπ s 2s ℓs/m

 

m X j=1

for some C > 0 that depends upon m only, and we see that ∞ X ℓ=1

However, ∞ X ℓ=1

|fbj (qφ(ℓ),ℓ + α)|2 

ℓ2s/m |fbφ (qφ(ℓ),ℓ + α)|2 < ∞.

ℓ2s/m |fbφ (qφ(ℓ),ℓ + α)|2 =

1/2

∞ ∞ X X ℓ2s/m 1 = ∞. = 1+2s/m ℓ ℓ ℓ=1 ℓ=1

(6.13)

(6.14)

The contradiction between (6.13) and (6.14) shows that fφ cannot be put in the form (6.9). Now, let P φ1 , φ2 , . . . , φr ∈ Φ be distinct, and consider a linear combination h = rj=1 dj fφj where, say, d1 6= 0. It follows from the observation (6.12) above that there is k0 ∈ N such that for all k > k0 with fbφ1 (k) 6= 0, we have fbφj (k) = 0 for all j ∈ {2, 3, . . . , r}. Thus, for all k > k0 with h(k) = d1 fbφ (k). Then, if we apply (6.13) and fbφ (k) 6= 0 , we see that b 1

1

(6.14) with fφ1 in place of fφ , we see that the contradiction between (6.13) and (6.14) applies to h, and we deduce that h cannot be written in the form (6.9). We now see that if V is the subspace of L2 ([0, 2π]) finitely spanned by {fφ : φ ∈ Φ}, then V has the required properties. 

7

Results for compact, connected abelian groups

In this section we look at some applications of the earlier results to compact, connected abelian groups and to the automatic continuity of linear forms on L2 spaces on these groups. Definitions. If z ∈ T and ν ∈ R we may write z uniquely as z = eit where t ∈ [0, 2π), and we then take z ν to be eitν . In particular, for α ∈ Z,

20

,

Rodney Nillsen

Generalised differences and multiplier operators

z α/2 is eitα/2 . Let α, β ∈ Z and let L be a linear form on L2 (T). Then L is called (α, β)-invariant if, for all b ∈ T and f ∈ L2 (T), h  i
L b(α+β)/2 δb + b−(α+β)/2 δb−1 ∗ f = b(α−β)/2 + b−(α−β)/2 L(f ).

Also, L is called translation invariant if L(δb ∗ f ) = L(f ) for all b ∈ G and f ∈ L2 (T).

Theorem 7.1 Let α, β ∈ Z and let L be a linear form on L2 (T). Then the following hold. (i) If L is (α, β)-invariant on L2 (T), then L is continuous on L2 (T). R Thus, in this case there is a function h ∈ L2 (T) such that L(f ) = T f h dµT for all f ∈ L2 (T. R (ii) If h ∈ L2 (T), the linear form on h ∈ L2 (T) given by f 7−→ T f h dµT is (α, β)-invariant if and only if there are c1 , c2 ∈ C such that h(z) = c1 z α + c2 z β for almost all z ∈ T. (iii) If L is a translation invariant linear form on L2 (T), L is a multiple of the Haar measure on T. Proof. (i) It follows immediately from Theorem 2.3 with s = 1 that if L is (α, β)-invariant, L vanishes on the space  f : f ∈ L2 (T) and fb(α) = fb(β) = 0 .

Thus, if L is (α, β)-invariant it vanishes on this closed subspace of L2 (T), a space that has finite codimension in L2 (T). By (d) of Proposition 5.1 in [9, page 25], L is continuous. The existence of the function h in this case comes from the fact that the dual space of L2 (T) is identified with L2 (T). (ii) If h ∈ L2 (T), by Theorem R 2.3 the linear form corresponding to h is (α, β)-invariant if and only if T f h dµT = 0 whenever fb(α) = fb(β) = 0. This occurs when b h(n) = 0 for all n ∈ Z with n 6= α, β. But that means that the Fourier expansion of h in L2 (T) is a linear combination of z α and zβ . (iii) If L is translation invariant, we see that it is (0, 0) invariant. Then, (ii) shows that L is a multiple of the Haar measure on T.  Note that the conclusion (iii) in Theorem 7.1 is due to Meisters and Schmidt [5]. Conclusion (ii) generalises their result. b Let G denote a compact, connected abelian group with dual group G. b The identity element in G is denoted by eb. The group operation in such 21

Rodney Nillsen

Generalised differences and multiplier operators

a group will be written multiplicatively, and the normalised Haar measure on such a group G will be denoted by µG . We denote by M (G) the family b of bounded complex Borel measures on G. Let m ∈ N and for each γ ∈ G m m with γ 6= eb let hγ : G −→ T be the function given by hγ (g1 , g2 , . . . , gm ) = (γ(g1 ), γ(g2 ), . . . , γ(gm )).

The function hγ is continuous and, as γ 6= eb, hγ maps Gm onto a compact connected subgroup of Tm that is strictly larger than {1}m , so this connected subgroup must be Tm itself, as Tm is connected. Consequently, hγ maps Gm onto Tm . It follows that for any non-negative measurable function on Tm , we have Z Z f dµT . (7.1) f ◦ hγ dµGm = Tm

Gm

This is because each side of (7.1) defines a translation invariant integral over Tm , so the equality in (7.1) is a consequence of the uniqueness of the Haar measure on T, mentioned in (iii) of Theorem 7.1. In the following result, note that for any compact connected abelian group G, for every element b ∈ G we have b = d2 for some d ∈ G [3, vol. I, page 385]. Theorem 7.2 Let G be a compact connected abelian group with dual group b Let eb be the identity element of G. b Let s ∈ N and let n, α, β ∈ Z with G. n∈ / {α, β}. Then, for a function f ∈ L2 (G) the following conditions (i) and (ii) are equivalent. (i) The Fourier transform of f vanishes at eb. That is, fb(b e) = 0. (ii) There are b1 , b2 , . . . , b4s+1 ∈ G and d1 , d2 , . . . , d4s+1 ∈ G with d2j = bj for all j ∈ {1, 2, . . . , 4s + 1}, such that there are f1 , f2 , . . . , f4s+1 ∈ L2 (G) so that f=

4s+1 X j=1

δdα−β + δd−(α−β) − δd2n−(α+β) − δd−(2n−(α+β)) j

j

j

j

s

∗ fj .

(7.2)

When f satisfies conditions (i) and (ii), almost all (b1 , b2 , . . . , b4s+1 ) ∈ G4s+1 have the following property: for any d1 , d2 , . . . , d4s+1 ∈ G with bj = d2j for all j ∈ {1, 2, . . . , 4s + 1}, there are f1 , f2 , . . . , f4s+1 ∈ L2 (G) such that (7.2) holds.

22

Rodney Nillsen

Generalised differences and multiplier operators

In the case when fb(b e) = 0 and α−β is even, for almost all (b1 , . . . , b4s+1 ) ∈ there are f1 , f2 , . . . , f4s+1 ∈ L2 (G) such that

G4s+1 , f=

4s+1 X

(α−β)/2

δbj

−(α−β)/2

+ δbj

j=1

n−(α+β)/2

− δbj

 −(n−(α+β)/2)) s

− δbj

∗ fj . (7.3)

Proof. Assume that (i) holds. In (7.1), given s ∈ N and n, α, β ∈ Z such that n 6= α and n 6= β, let’s take f to be the function on T4s+1 whose value f (z1 , z2 , . . . , z4s+1 ) at (z1 , z2 , . . . , z4s+1 ) is 1 4s+1 X j=1

(α−β)/2 −(α−β)/2 n−(α+β)/2 −(n−(α+β))/2 2s + zj − zj − zj zj

.

b with γ 6= eb we have Then, using (7.1) with m = 4s + 1, for each γ ∈ G Z dµG (b1 )dµG (b2 ) · · · dµG (b4s+1 ) 4s+1 2s G4s+1 X (α−β)/2 −(α−β)/2 n−(α+β)/2 −(n−(α+β)/2) ) + γ(b ) − γ(b ) − γ(b ) γ(bj j j j j=1

1 = 6s+1 4s+1 2 π

Z

[0,2π]4s+1

4s+1 X j=1

cos



dx1 dx2 · · · dx4s+1     2s .   α−β α+β xj − cos xj n− 2 2 (7.4)

b Then, γ(d)2 = γ(b), Now let b, d ∈ G with d2 = b, and let γ ∈ G. iθ so that if we put γ(b) = e where θ ∈ [0, 2π), we have γ(d) = eiθ/2 or γ(d) = ei(θ/2+π) . In the former case we have γ(d)α−β = eiθ(α−β)/2 = γ(b)(α−β)/2 ,

(7.5)

while in the latter case we have γ(d)α−β = ei(α−β)π eiθ(α−β)/2 = (−1)α−β γ(b)(α−β)/2 .

(7.6)

Similarly, when γ(d) = eiθ/2 or γ(d) = ei(θ/2+π) we have, respectively, γ(d)α+β = γ(b)(α+β)/2 or γ(d)α+β = (−1)α+β γ(b)(α+β)/2 .

23

(7.7)

Rodney Nillsen

Generalised differences and multiplier operators

Note that in (7.5), (7.6) and (7.7), α − β and α + β are both even or both odd, so (−1)α−β and (−1)α+β are both equal to 1 or both equal to −1. Now, for n ∈ Z and b, d ∈ G with d2 = b, put
s λb,d,n = δdα−β + δd−(α−β) − δd2n−(α+β) − δd−(2n−(α+β)) ∈ M (G). b Then, for γ ∈ G,


bb,d,n (γ) = γ(d)−(α−β) + γ(d)α−β − γ(d)−(2n−(α+β)) − γ(d)2n−α−β s . (7.8) λ

In view of (7.5), (7.6) and (7.7), we see that bb,d,n (γ)| = γ(b)(α−β)/2 + γ(b)−(α−β)/2 − γ(b)n−(α+β)/2 − γ(b)−(n−(α+β)/2) . |λ (7.9) Now, for each b ∈ G, let db ∈ G be any element such that d2b = b. As n ∈ / {α, β}, if M is the constant as in Lemma 2.2 and we use (7.4) and (7.9), upon changing the order of summation and integration we have   Z

  m Y  X X M |fb(γ)|2   dµ (b ) ≤ |fb(γ)|2 ,   G j 6s+1 4s+1 4s+1   2 π 4s+1 X G b b =eb γ∈G,γ6 γ∈G bb ,d ,n (γ)|2  j=1 |λ j bj j=1

(7.10)

which is finite by Plancherel’s Theorem (see [3, vol. II, page 226]). We deduce that provided fb(b e) = 0, for almost all (b1 , b2 , . . . , b4s+1 ) ∈ G4s+1 we have that X |fb(γ)|2 < ∞. (7.11) 4s+1 X b γ∈G bb ,d ,n (γ)|2 |λ j bj j=1

Then, (ii) follows from (7.8), (7.11) and Theorem 2.1, so (i) implies (ii). It is clear from (7.8) that if f has the form (7.2), then fb(b e) = 0. Thus, (ii) implies (i). Above, the observation was made that (7.11) holds for almost all (b1 , b2 , . . . , b4s+1 ) ∈ G4s+1 . If (b1 , b2 , . . . , b4s+1 ) is any such point, we deduce that for any choice of elements d1 , d2 , . . . , d4s+1 such that d2j = bj for all j, and when fb(b e) = 0, we will have (7.2) holding. 24

Rodney Nillsen

Generalised differences and multiplier operators

When fb(b e) = 0 and α − β is even, the final conclusion derives from the above arguments and the fact that (α−β)/2

dα−β = (d2j )(α−β)/2 = bj j

(α+β)/2

and dα+β = bj j

.

 The following is a result concerning automatic continuity on groups. It is derived from Theorem 7.2, but only a special case is stated. More general results can be derived from Theorem 7.2. Theorem 7.3 Let G be a compact connected abelian group. Then the following conditions (i), (ii) and (iii) on a linear form L : L2 (G) −→ C are equivalent. (i) L is translation invariant. That is, L(δg ∗ f ) = L(f ), for all g ∈ G and f ∈ L2 (G). (ii) There is n ∈ Z with n ∈ / {−1, 1} such that

L (δg + δg−1 ) ∗ f = L (δgn + δg−n ) ∗ f ,

for all g ∈ G and f ∈ L2 (G). (iii) L is a multiple of the Haar measure. Also, the normalised Haar measure µG on G is unique.

Proof. (i) implies that (ii) holds for any n ∈ N, so it must hold for any particular n. Also (iii) implies (i) because the Haar measure is translation invariant. Finally, assume that (ii) holds for some n ∈ Z with n ∈ / {−1, 1}. By Theorem 7.2 with s = 1, α = 1 and β = −1, we deduce that L vanishes on the closed subspace {f : f ∈ L2 (G) and fb(b e) = 0}. This latter space has 2 codimension 1 in L (G) and so it follows easily that L is continuous and is a multiple of the Haar measure on G (see [5, page 415]), and (iii) follows. Finally, as the Haar measure µG defines a translation invariant linear form on L2 (G), the equivalence of (i) and (iii) implies the uniqueness of the Haar measure. 

References [1] J. Bourgain, Translation invariant forms on Lp (G), 1 < p < ∞, Ann. Inst. Fourier (Grenoble), 36 (1986), 97-104. [2] R. E. Edwards, Fourier series, volume 1, Holt, Rinehart and Winston, New York, 1967. 25

Rodney Nillsen

Generalised differences and multiplier operators

[3] E. Hewitt and K. A. Ross, Abstract Harmonic Analysis, Volume I and Volume II, Springer -Verlag, New York 1963 and 1970. [4] B. E. Johnson, A proof of the translation invariant form conjecture for L2 (G), Bull. des Sciences Math. 107(1983), 301-310. [5] G. Meisters and W. Schmidt, Translation invariant linear forms on L2 (G) for compact abelian groups G, Journ. Func. Anal. 11 (1972), 407-424. [6] G. Meisters. Some discontinuous translation-invariant linear forms, Journ. Func. Anal. 12 (1973), 199-210 [7] G. Meisters. Some problems and results on translation-invariant linear forms, Lecture Notes in Mathematics, vol. 975 (J. M. Bachar, W. G Bade, et al, eds.), 423-444, Springer-Verlag, New York, 1983. [8] R. Nillsen, Banach spaces of functions and distributions characterized by singular integrals involving the Fourier transform, Journ. Func. Anal. 110 (1992), 73-95. [9] R. Nillsen, Difference spaces and Invariant Linear Forms, Lecture Notes in Mathematics 1586, Springer-Verlag, Berlin Heidelberg New York, 1994. [10] R. Nillsen, The Fourier transform and finite differences: an exposition and proof of a result of Gary H. Meisters and Wolfgang M. Schmidt, at h http://www.uow.edu.au/∼nillsen/differences on circle.html i, University of Wollongong, June, 2013. [11] I. Niven, Irrational Numbers, Carus Mathematical Monographs 11, Mathematical Association of America, 1967. [12] K. A. Ross, A trip from classical to abstract Fourier analysis, Notices Amer. Math. Soc. 61 (2014), 1032-1038. [13] W. H. Schmidt, Diophantine Approximation, Lecture Notes in Mathematics 785, Springer-Verlag, Berlin Heidelberg New York, 1980. [14] K. R. Stromberg, An Introduction to Classical Real Analysis, Wadsworth, Belmont, 1981.

26

Rodney Nillsen

Generalised differences and multiplier operators

Rodney Nillsen School of Mathematics and Applied Statistics University of Wollongong New South Wales AUSTRALIA 2522 email: [email protected] web page: http://www.uow.edu.au/∼nillsen

27